| 29th March 2013 05:45 PM | |
| Arvind Kumar | Re: CBSE Test Paper For Class 11 you are looking for the CBSE 11th class question paper of Maths, i am giving here: 1. Write the following set in the set-builder form: {5, 25, 125, 625}. Sol. It can be seen that 5 = 51, 25 = 52, 125 = 53, and 625 = 54. ∴ {5, 25, 125, 625} = {x: x = 5n, n ∈ N and 1 ≤ n ≤ 4} 2. Let A and B be two sets such that n(A) = 3 and n (B) = 2. If (x, 1), (y, 2), (z, 1) are in A × B, find A and B, where x, y and z are distinct elements. Sol.. It is given that n(A) = 3 and n(B) = 2; and (x, 1), (y, 2), (z, 1) are in A × B. We know that A = Set of first elements of the ordered pair elements of A × B B = Set of second elements of the ordered pair elements of A × B. ∴ x, y, and z are the elements of A; and 1 and 2 are the elements of B. Since n(A) = 3 and n(B) = 2, it is clear that A = {x, y, z} and B = {1, 2}. 3. Let f be the subset of Z ×Z defined by f={(ab,a+b):a,b∈Z} Is f a function from Z to Z? Justify your answer. Sol. We observe that 1 x 6 = 6 and 2 x 3 = 6 ⇒ (1×6,1+6) ∈ f and (2×3,2+3) ∈ f ⇒ (6,7) ∈ f and (6,5) ∈ f. We observe that an element 6 have appeared more than once as the first component of the ordered pairs in f. So f is not a function for Z to Z. 4. Express the given complex number in the form a + ib: (1 – i) – (–1 + i6). Sol. (1 – i) – (–1 + i6) = 1 – i + 1 – 6i = 2 – 7i 5. If 2x+ i (x - y) = 5, where x and y are real numbers, find the values of x and y. Sol. We have 2x+ i(x - y) = 5 or 2x+ i(x - y) = 5 + 0.i Comparing the real and imaginary parts,we get 2x = 5 and x - y = 0 ⇒ x = 5/2 and x = y Thus x = y = 5/2. 6. Solve the given inequality for real x: 4x + 3 < 5x + 7. Sol. 4x + 3 < 5x + 7 ⇒ 4x + 3 – 7 < 5x + 7 – 7 ⇒ 4x – 4 < 5x ⇒ 4x – 4 – 4x < 5x – 4x ⇒ –4 < x Thus, all real numbers x, which are greater than –4, are the solutions of the given inequality. Hence, the solution set of the given inequality is (–4, ∞). 7. How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated? Sol. There will be as many ways as there are ways of filling 3 vacant places In succession by the given six digits. In this case, the units place can be filled by 2 or 4 or 6 only i.e., the units place can be filled in 3 ways. The tens place can be filled by any of the 6 digits in 6 different ways and also the hundreds place can be filled by any of the 6 digits in 6 different ways, as the digits can be repeated. Therefore, by multiplication principle, the required number of three digit even numbers is 3 × 6 × 6 = 108. 8. Find the sum of odd integer from 1 to 21. Sol. The odd integer from 1 to 21 are 11, namely 1, 3, 5, 7, 9, 11,13,15,17,19, 21. 9. Write the equations for the x and y-axes. Sol. The y-coordinate of every point on the x-axis is 0. Therefore, the equation of the x-axis is y = 0. The x-coordinate of every point on the y-axis is 0. Therefore, the equation of the y-axis is y = 0. 10. A die is rolled. Let E be the event “die shows 4” and F be the event “die shows even number”. Are E and F mutually exclusive? Sol. When a die is rolled, the sample space is given by S = {1, 2, 3, 4, 5, 6} Accordingly, E = {4} and F = {2, 4, 6} It is observed that E ∩ F = {4} ≠ Φ Therefore, E and F are not mutually exclusive events. 11. A wheel makes 500 revolutions per minute. How many radians it turns in one second? Sol. In one revolution angle covered is 2 π radians In one minute, number of revolutions = 500 In one second, number of revolutions = 500/60 = 50/6 ∴ In one second angle covered = 50/6 X 2 π radians 12. Write the value of tan 75°. Sol. tan 75° = tan(45°+30°) 13. In a survey of 600 students in a school, 150 students were found to be taking tea and 225 taking coffee, 100 were taking both tea and coffee. Find how many students were taking neither tea nor coffee? Sol. Let U be the set of all students who took part in the survey. Let T be the set of students taking tea. Let C be the set of students taking coffee. Accordingly, n(U) = 600, n(T) = 150, n(C) = 225, n(T ∩ C) = 100 13. In a survey of 600 students in a school, 150 students were found to be taking tea and 225 taking coffee, 100 were taking both tea and coffee. Find how many students were taking neither tea nor coffee? Sol. Let U be the set of all students who took part in the survey. Let T be the set of students taking tea. Let C be the set of students taking coffee. Accordingly, n(U) = 600, n(T) = 150, n(C) = 225, n(T ∩ C) = 100 To find: Number of student taking neither tea nor coffee i.e., we have to find n(T' ∩ C'). n(T' ∩ C') = n(T ∪ C)' = n(U) – n(T ∪ C) = n(U) – [n(T) + n(C) – n(T ∩ C)] = 600 – [150 + 225 – 100] = 600 – 275 = 325 Hence, 325 students were taking neither tea nor coffee. For more detailed question paper, here is attachment word file (CBSE Class 11th Maths Question paper) |
| 29th March 2013 02:55 PM | |
| pooja Jaju | CBSE Test Paper For Class 11 Will you please provide me the CBSE CBSE Test Paper For Class 11 Test paper of Maths? |
